A ferry can carry a maximum of 20 adults or 50 children. A ticket for an adult cost $50 and a ticket for a child cost $25. A group of adults and children took 3 such ferries at maximum load on a ferry trip. The total amount paid for the children was $750 less than the total amount paid for the adults.
- How many adults were there?
- How many children were there?
Number of adults |
Amount paid for adults |
Number of children |
Amount paid for children |
Difference in total amounts paid |
60 |
60 x 50 = 3000 |
0 |
0 x 25 = 0 |
3000 - 0 = 3000 |
- 2 |
|
+ 5 |
|
- 225 |
58 |
58 x 50 = 2900 |
5 |
5 x 25 = 125 |
2900 - 125 = 2775 |
40 |
|
50 |
|
750 |
20 adults → 50 children
(÷10)2 adults → 5 children
From every decrease of 2 adults, there is an increase of 5 children and the difference in the total amounts paid between the adults and the children will be reduced by $225.
If all 3 ferries are occupied by only adults, total number of adults
= 3 x 20
= 60
Total difference
= 3000 - 750
= $2250
Small difference
= 3000 - 2775
= $225
Number of sets
= 2250 ÷ 225
= 10
Total decrease in the number of adults
= 10 x 2
= 20
(a)
Number of adults
= 60 - 20
= 40
(b)
Number of children
= 10 x 5
= 50
Answer(s): (a) 40; (b) 50